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The figure given shows two identical capacitors $C_1$ and $C_2$, each of $2 \mu F$ capacitance, connected to a battery of $5\; V$. Initially switch 'S' is closed. After some times is left open and dielectric slabs of dielectric constant K = 5 are inserted to fill completely the space between the plates of the two capacitors. How will the potential difference between the plates of the capacitors be affected after the slabs are inserted ?

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$(B)\; \large\frac{1}{K}$
Hence B is the correct answer.
answered Jun 16, 2014 by meena.p

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